Please explain as much as possible! I really struggle with these, especially oxi

Question

Please explain as much as possible! I really struggle with these, especially oxidation states! Thanks

A chemist designs a galvanic cell that uses these two half-reactions: half-reactiorn standard reduction potential NO3(aq)+4H+(aq)+3e- NO(g)+2H20( E red = +0.96 V Fe3+ (aq)+e- Fe2+ (aq) red Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode. Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written Do you have enough information to calculate the cell voltage under standard conditions? Yes No If you said it was possible to calculate the cell voltage, do so and enter your answer V here. Round your answer to significant digits

Explanation / Answer

a)

cathode --> species which reduces, note that this is the HIGHEST potential

from thel ist

0.771 < 0.96

then

NO3-(aq) + 4H+(aq) + 3e- --> NO(g) + 2H2O(l)

b)

the anode contains oxidation, which is the Fe2+ to Fe+3 reaction

Fe+2(aq) --> Fe3+(aq) + e-  

balance

3Fe+2(aq) --> 3Fe3+(aq) + 3e-  

c)

add both

NO3-(aq) + 4H+(aq) + 3e- --> NO(g) + 2H2O(l)

3Fe+2(aq) --> 3Fe3+(aq) + 3e-  

3Fe+2(aq) + NO3-(aq) + 4H+(aq) + 3e- --> NO(g) + 2H2O(l) + 3Fe3+(aq) + 3e-  

add all

3Fe+2(aq) + NO3-(aq) + 4H+(aq) --> NO(g) + 2H2O(l) + 3Fe3+(aq)

d) and e)

yes, we have enough info, both E|cell data

E°cell = Ecathode - Eanode = 0.96 - 0.771 = 0.189 V

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